Anyone here into math?

[Guest [ou...]]

Answer to the 0 question.

 

By definition, a number is even if it is an integer multiple of 2.

Since 0=2*0 zero is an even number.

 

Answer to the

 

Russian roulette problem

 

Assumptions & Rules:

1. There are 2 players.

2. The revolver contains 1 bullet.

3. The cylinder consists of 5 or 6 chambers (but one could generalize to N).

4. Each player takes a shot at the opponent in turn.

 

Variants:

1. The game ends after 2 turns, possibly resulting in a draw.

2. The game does not end until we have a winner and the cylinder is

either A. spun after each turn or B. not.

 

Question: Which player should go 1st in each variant?

 

 

Variant 1.

 

Let P(A) and P(B) be the probabilities that the 1st or 2nd player wins the game.

 

P(B) = P(B-A) = P(A') * P(B|A') = (N-1)/N * 1/(N-1) = 1/N = P(A)

 

Hence it doesn't matter who goes 1st and the game ends in a draw with probability:

 

P = 1 - P(A) - P(B) = 1 - 2/N = (N-2)/N

 

For N=5 each player has a 20% chance of winning with the game ending in a draw 60% of the time.

For N=6 each player has a 1/6 chance of winning with the game ending in a draw 2/3 of the time.

 

 

Variant 2a.

 

Let P(A) and P(B) be the probabilities that the 1st or 2nd player wins the game.

 

P(B) is equal to the probability that B survives A's first shot

times the probability that they win the game from then on.

If B survives the 1st round, they turn into the player to go 1st.

Combining this with the fact that P(AUB)=1 gives:

 

P(A) + P(B) = 1  &  P(B) = (N-1)/N * P(A)

 

Putting those together gives:

 

P(A) = N/(2N-1)  &  P(B) = (N-1)/(2N-1)

 

Hence player A has a greater chance of winning.

For N=(5 or 6) we get P(A)=(5/9 or 6/11)=55% and P(B)=(4/9 or 5/11)=45%

 

 

Variant 2b.

 

Let P(N) be the probability that the 1st player wins the game given N unfired rounds.

 

For the 1st player to win, they have to either win the game in the 1st round

or survive the 2st round and win the game from then on. Thus:

 

P(N) = 1/N + (N-1)/N * (N-2)/(N-1) * P(N-2) = 1/N + (N-2)/N * P(N-2)

 

Obviously P(1)=1 and P(2)=1/2. We can also observe that:

 

P(N) >= N/2 * 1/N = 1/2

 

By induction we get that P(N)=1/2 for evens and P(N)>1/2 otherwise, confirming that P(N)>=1/2=50%

Hence the 1st player has a greater chance of winning.

For N=5 we get P(5)=60%

[Guest [Kp...]]

Answer to the 0 question.

 

By definition, a number is even if it is an integer multiple of 2.

Since 0=2*0 zero is an even number.

 

i don't know why you asked this question. wasn't the answer obvious? i'm sure you had another purpose behind it -- was there one?

 

Answer to the

 

Russian roulette problem

 

Assumptions & Rules:

1. There are 2 players.

2. The revolver contains 1 bullet.

3. The cylinder consists of 5 or 6 chambers (but one could generalize to N).

4. Each player takes a shot at the opponent in turn.

 

Variants:

1. The game ends after 2 turns, possibly resulting in a draw.

2. The game does not end until we have a winner and the cylinder is

either A. spun after each turn or B. not.

 

Question: Which player should go 1st in each variant?

 

 

Variant 1.

 

Let P(A) and P(B) be the probabilities that the 1st or 2nd player wins the game.

 

P(B) = P(B-A) = P(A') * P(B|A') = (N-1)/N * 1/(N-1) = 1/N = P(A)

 

Hence it doesn't matter who goes 1st and the game ends in a draw with probability:

 

P = 1 - P(A) - P(B) = 1 - 2/N = (N-2)/N

 

For N=5 each player has a 20% chance of winning with the game ending in a draw 60% of the time.

For N=6 each player has a 1/6 chance of winning with the game ending in a draw 2/3 of the time.

 

 

Variant 2a.

 

Let P(A) and P(B) be the probabilities that the 1st or 2nd player wins the game.

 

P(B) is equal to the probability that B survives A's first shot

times the probability that they win the game from then on.

If B survives the 1st round, they turn into the player to go 1st.

Combining this with the fact that P(AUB)=1 gives:

 

P(A) + P(B) = 1  &  P(B) = (N-1)/N * P(A)

 

Putting those together gives:

 

P(A) = N/(2N-1)  &  P(B) = (N-1)/(2N-1)

 

Hence player A has a greater chance of winning.

For N=(5 or 6) we get P(A)=(5/9 or 6/11)=55% and P(B)=(4/9 or 5/11)=45%

 

 

Variant 2b.

 

Let P(N) be the probability that the 1st player wins the game given N unfired rounds.

 

For the 1st player to win, they have to either win the game in the 1st round

or survive the 2st round and win the game from then on. Thus:

 

P(N) = 1/N + (N-1)/N * (N-2)/(N-1) * P(N-2) = 1/N + (N-2)/N * P(N-2)

 

Obviously P(1)=1 and P(2)=1/2. We can also observe that:

 

P(N) >= N/2 * 1/N = 1/2

 

By induction we get that P(N)=1/2 for evens and P(N)>1/2 otherwise, confirming that P(N)>=1/2=50%

Hence the 1st player has a greater chance of winning.

For N=5 we get P(5)=60%

 

very nice, very elegant!

 

------------

 

my head is swimming in a sea of probabilities, godel and nature. i've read a bit during this time and i've discovered a few facets about numbers (that i did not know earlier) and realized that man discovered these facets in 500 CE or earlier.  nothing that i think and come up with has not been discovered or invented before (and which does not speak as poorly of me as it seems). 

 

numbers aside, and even though i wasn't a physics student, i think, after the reading i did in the last few days, i can say that i belong to the penrose school of thought -- string theory may not be true and QM works but it also may not be an entirely satisfactory interpretation of nature (i think this is what penrose said about QM).

 

edit. and i read this essay too (it is very dense and packed with information and opinions in every sentence and needed multiple readings).

[Guest [ou...]]

1. Just an easy question to lure in more members since some believe 0 is neither even nor odd.

2. What do you think of the x^x^x^...=2 equation I posted?

3. What do you think of my answer to chessplayer's problem?

4. What's your opinion on future problems?

a. Should I stick to probability?

b. Should I add more algebra?

c. What about number theory or calculus?

d. Would you be interested in physics (mainly classical mechanics) problems?

[Guest [Kp...]]

2. What do you think of the x^x^x^...=2 equation I posted?

i don't remember it. i think i must have ignored it. honestly, anything beyond high school math intimidates me (at this age).

3. What do you think of my answer to chessplayer's problem?

it was nice of you to return and revise your answer. i liked that. you might have googled or mightn't but what was more important was that you corrected your mistake when it came to light on a subsequent rethink -- it's a hallmark of humility. i could not solve the problem and was waiting for your answer. you were thinking about your answer even after having solved it for something in you was not satisfied -- that is a quest for perfection.

4. What's your opinion on future problems?

i can speak only for myself. i do not know if others here are interested in this section. however, it would be a good idea if you reached out to leslieash to find out what sort of problems she would enjoy. if they are coding-math problems (like say discrete math) then even i might give it a shot. ditto for chessplayer and badsocref and magnesi (did i leave anyone out?). i think more people would be interested in problems that require tricky thinking than a high level of sophistication or calculus or advanced probability (btw chessplayer's puzzle was not an "advanced" probability question -- advanced would be ones requiring use of formulae -- but it was tough).

a. Should I stick to probability?

you can and you probably should. (HH vs HT was slightly tough though -- but still OK.)

b. Should I add more algebra?

100%. algebra is what most people love i think. "please stop asking us to find your x. she is not coming back and don't ask y."

c. What about number theory or calculus?

i suggest you reach out to badsocref and others. it all depends on how comfortable they are with it. i'm ok with simple derivation, integration and trig. but series and sums- NO!

d. Would you be interested in physics (mainly classical mechanics) problems?

yes.

 

maybe reaching out publicly and privately to above members would be a good idea? anyway, your call (it matters little either way for this is a benzo forum and not ask dr. math!).

 

[Guest [ou...]]

Thanks for your input.

I think I'll be posting 2 problems at a time; one easy but tricky and a hard one.

 

1. (chessplayer question was not an "advanced" probability question -- advanced would be ones requiring use of formulae)

2. "please stop asking us to find your x. she is not coming back and don't ask y."

3. reaching out ... would be a good idea ... (it matters little ... for this is a benzo forum)

 

1. When I corrected my answer, I did so by trying to calculate the exact frequency,

the formulae for which are pretty advanced.

 

2. PRICELESS!!

 

3. I know - that's why it's OFF-Topic - but taking your mind off of withdrawal is a good coping strategy.

As for LeslieAsh in particular, I think she IS interested but being a moderator takes LOTS of time.

[Guest [Kp...]]

by the bye -- i have not been tapering since 13th october (diwali-festival break) and i am glad to report that the sleep-onset insomnia has resolved. so, yes, it was withdrawal related and it will probably recur when i resume taper on the 21st of october.

[Guest [ou...]]

In my latest post I wrote that I'll be posting 2 problems at a time; 1 easy and 1 hard.

However, I came up with several problems in that range;

mostly algebraic (with a bit of calculus for good measure) since we've had enough of probability.

I'm posting them all at once because... why not?

Also, I'm using asterisks to specify the difficulty of each one.

Also also, I'll be giving hints to anyone interested.

 

 

Problems:

 

 

1. Solve the following equations:

 

A. 5x^2 - 4xy + y^2 - 2x + 1 = 0  !Solved

 

B. 9^lnx - 2*x^ln3 = 3

 

C. x^4 - 6x^3 + 12x^2 -12x + 4 = 0

 

2. Simplify X = ((50)^(1/2) + 7)^(1/3) - ((50)^(1/2) - 7)^(1/3)

 

3. Given an increasing function such that f(f(x))=x show that f(x)=x

 

4. Given a function such that f(f(x))=4x-3 and f(f(f(x)))=8x+c show that f(x)=2x-1

 

5. Show that a function satisfying the inequality |f(x)-f(y)|<=(x-y)^2 is constant.

 

6. Given a positive function such that f(x+y)=f(x)*f(y) and f'(0)=1 show that f(x)=e^x

 

 

Have fun!

 

 

Edit:

a. I'm rescinding Problems 5 & 6 since they are too hard for the purpose of this thread.

b. I'm also rescinding the asterisks since the rest of the problems are about equally difficult.

[Guest [ba...]]

1C)  0.58578644

[Guest [ba...]]

1B) e  (2.718281828)

[Guest [ou...]]

1B) e  (2.718281828)

1C)  0.58578644

 

1B) Correct, though I suspect you solved it by eye.

1C) Less than half-right since it has 2 solutions and,

unlike e you gave a ballpark figure so I suspect you cheated.

[Guest [Kp...]]

*3. Given an increasing function such that f(f(x))=x show that f(x)=x

 

when f'(x) > 0, f(f(x)) = x is true for f(x) = x on substitution

thus f(x) = x

 

(i think this was the shortest proof of my life)

[Guest [Kp...]]

...so I suspect you cheated.

 

all exams are open book unless specified otherwise.

but seriously, you can ask for reasoning. it serves no purpose calling names.

[Guest [Kp...]]

*4. Given a function such that f(f(x))=4x-3 and f(f(f(x)))=8x+c show that f(x)=2x-1

 

assume f(x) = 2x - 1

so f(f(x)) = 2f(x) - 1

so f(f(x)) = 2 * (2x - 1) - 1 by substitution of value of f(x)

so f(f(x) = 4x - 3 (which is equation 1, the given)

 

f(f(x)) = 4x - 3

so f(f(fx))) = 4(f(f(x)) - 3

so f(f(f(x))) = 4(4x -3) - 3

so f(f(f(x))) = 16x - 15 <> equation 2, the given (? ? ?)

 

something is wrong?

[Guest [Kp...]]

A. 5x^2 - 4xy + y^2 - 2x + 1 = 0

 

5x² - 4xy + y ² - 2x + 1 = 0

(5x² - 2x + 1) - 4xy + y² = 0

 

can't factor the quadratic equation... actually no clue!

 

[Guest [ou...]]

Given an increasing function such that f(f(x))=x show that f(x)=x

 

when f'(x) > 0, f(f(x)) = x is true for f(x) = x on substitution

thus f(x) = x

 

I didn't say f was differentiable. Nice try though.

 

all exams are open book unless specified otherwise

 

book <> calculator

 

Given a function such that f(f(x))=4x-3 and f(f(f(x)))=8x+c show that f(x)=2x-1

 

assume f(x) = 2x - 1

so f(f(x) = 4x - 3

so f(f(fx))) = 4(f(f(x)) - 3 = 16x - 15 <> equation 2, the given (? ? ?)

 

f(f(fx))) = 4f(x)-3 or 2f(f(x))-1 = 8x-7

 

5x² - 4xy + y² - 2x + 1 = 0

(5x² - 2x + 1) - 4xy + y² = 0

 

can't factor the quadratic equation...

 

Wrong factoring choice.

[Guest [Kp...]]

5x² - 4xy + y² - 2x + 1 = 0

(5x² - 2x + 1) - 4xy + y² = 0

 

can't factor the quadratic equation...

 

Wrong factoring choice.

 

ok, there are 2 polynomials

 

5x² - 4xy + y² - 2x + 1 = 0

so 4x² - 4xy +y² +x² - 2x + 1 = 0

so (2x - y)² + (x - 1)² = 0

since square of a number is positive, (2x - y) = 0 and (x - 1) = 0

so  2x - y = x - 1

so x = y - 1

 

ans. x = y -1

 

[Guest [Kp...]]

Edit:

a. I'm receding Problems 5 & 6 since they are too hard for the purpose of this thread.

b. I'm also receding the asterisks since the rest of the problems are about equally difficult.

 

oh, i just noticed the edit.

 

rescinding.  (don't mind... i am a bit fastidious but i'm not a grammar nazi... there is a difference, lol)

[Guest [ou...]]

5x² - 4xy +y² - 2x + 1 = 0

so 4x² - 4xy +y² +x² - 2x + 1 = 0

so (2x - y)² + (x - 1)² = 0

since square of a number is positive, (2x - y) = 0 and (x - 1) = 0

so  2x - y = x - 1

so x = y - 1

 

ans. x = y -1

 

SO close but incomplete.

Combine your answer with the fact that x=1 and you get y=2.

Problem solved!

It was a system 'disguised' as an equation.

 

... I'm receding ...

 

rescinding (don't mind... i am a bit fastidious but i'm not a grammar nazi)

 

Since I AM a grammar nazi I do NOT mind.

In fact, now I'm forced to edit my post yet again.

[Guest [ou...]]

Problems:

 

1. Solve the following equations:

 

A. 5x^2 - 4xy + y^2 - 2x + 1 = 0

 

B. 9^lnx - 2*x^ln3 = 3

 

C. x^4 - 6x^3 + 12x^2 -12x + 4 = 0

 

2. Simplify X = ((50)^(1/2) + 7)^(1/3) - ((50)^(1/2) - 7)^(1/3)

 

3. Given an increasing function such that f(f(x))=x show that f(x)=x

 

4. Given a function such that f(f(x))=4x-3 and f(f(f(x)))=8x+c show that f(x)=2x-1

 

Hints:

 

A. Solved!

B. Solved!

C. Divide by x^2

2. Solved!

3. Assume the opposite

4. Observe that f[f(f(x))]=f(f[f(x)])

[Guest [Kp...]]

2. Simplify X = ((50)^(1/2) + 7)^(1/3) - ((50)^(1/2) - 7)^(1/3)

 

 

using (√2+1)^3 = 2√2+1+6+3√2 and (√2-1)^3 = 2√2-1-6+3√2

 

x = (√50+7)^(1/3) - (√50-7)^(1/3)

x = (5√2+7)^(1/3) - (5√2-7)^(1/3)

x = (2√2+1+6+3√2)^(1/3) - ( 2√2-1-6+3√2)^(1/3)

x = (√2+1)^(3/3)-(√2-1)^(3/3)

x = √2+1-√2+1

x = 2

 

ans. x = 2

[Guest [Kp...]]

B. 9^lnx - 2*x^ln3 = 3

 

since a^lnb=b^lna

 

9^lnx - 2*x^ln3 = 3

3*3^lnx - 2*x^ln3 = 3

3*x^ln3 - 2*x^ln3 = 3

x^ln3 = 3

since e^lnx = x

x = e

 

ans. x = e

[Guest [ou...]]

using (√2+1)^3 = 2√2+1+6+3√2 and (√2-1)^3 = 2√2-1-6+3√2

 

x = (√50+7)^(1/3) - (√50-7)^(1/3)

x = (5√2+7)^(1/3) - (5√2-7)^(1/3)

x = (2√2+1+6+3√2)^(1/3) - ( 2√2-1-6+3√2)^(1/3)

x = (√2+1)^(3/3)-(√2-1)^(3/3)

x = √2+1-√2+1

x = 2

 

 

9^lnx - 2*x^ln3 = 3

3*3^lnx - 2*x^ln3 = 3

3*x^ln3 - 2*x^ln3 = 3

x^ln3 = 3

since e^lnx = x

x = e

 

Both correct and elegant!

Marking the corresponding problems solved.

[Guest [ou...]]

Here's a list of the remaining problems (plus a new one) along with hints.

 

 

Problems:

 

1. Solve  x^4 - 6x^3 + 12x^2 -12x + 4 = 0

2. Given an increasing function such that f(f(x))=x show that f(x)=x

3. Given a function such that f(f(x))=4x-3 and f(f(f(x)))=8x+c show that f(x)=2x-1

4. Given a function such that f(x^2)-y^2=f(x+y)*f(x-y) show that f(x)=x

 

 

Hints:

 

1. Divide by x^2

2. Assume the opposite

3. c=-7

4. Set x=y

[Guest [ou...]]

9^lnx - 2*x^ln3 = 3

3*3^lnx - 2*x^ln3 = 3

3*x^ln3 - 2*x^ln3 = 3

x^ln3 = 3

since e^lnx = x

x = e

 

Both correct and elegant!

 

Actually, I missed a mistake as you arrived at the correct answer (by accident).

Here is the mistake and the correct derivation.

 

9^lnx=(3*3)^lnx=(3^2)^lnx=(3^lnx)^2=(x^ln3)^2<>3*x^ln3 for x<>e

 

9^lnx - 2*x^ln3 = 3

(3^lnx)^2 - 2*3^lnx - 3 = 0

3^lnx = 3 or -1<0

lnx = 1

x = e

[Guest [ou...]]

Here's a list of the remaining problems along with extra hints.

 

 

Problems:

 

1. Solve  x^4 - 6x^3 + 12x^2 -12x + 4 = 0

2. Given an increasing function such that f(f(x))=x show that f(x)=x

3. Given a function such that f(f(x))=4x-3 and f(f(f(x)))=8x+c show that f(x)=2x-1

4. Given a function such that f(x^2)-y^2=f(x+y)*f(x-y) show that f(x)=x

 

 

Hints & Tips:

 

1. Divide by x^2 and set y=x+2/x

2. Assume the opposite

3. c=-7

4. Set y=x and show that f(0)=0